A constant torque of 30m⋅N is applied to the rim of a 6.0-kg uniform disk of radius 0.20m .What is the angular speed of the disk about an axis through its center after it rotates 2.0 revolutions from rest?

Answer 1

v = a t

where a = alpha = angular acceleration
= torque/I

here I is moment of inertia of uniform disc = (1/2) m r^2

so I = (1/2)(6)(.04) = .12
and
a = 30/.12 = 250 radians/s^2

then v = 250 t
and theta = (1/2)a t^2 = 4 pi radians
so
125 t^2 = 4 pi
t = .317 second
then v = 250(.317) = 79.3 radians/second